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Proof step 6 from proof size-three-01 (previous step)

# s0 < 2, c ∈ s0, b ∈ s0 ⊦ b = c, a ∈ s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)

Rule argument

spec: <toplevel> name: indef seq: ⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0) subst: [s, a] → [s0, b] flags: <all paths>

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%size-zero-02: ⊦ # s = 0 → s = ∅;%ne-01: ⊦ a ∈ s → s ≠ ∅;%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; (s)% ⊦ m < 1 ↔ m = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: s0, b